A solid has a mass of 200g in air and when partly immersed in a certain liquid it has a mass of 160g. Given that the relative density of the liquid is 0.80, calculate the volume of the solid immersed in the liquid.
the displaced liquid has a mass of 40 g ... (200 g - 160 g)
volume = mass / density = 40 g / 0.80 = ? cm^3
Yes
To find the volume of the solid immersed in the liquid, we can use the concept of relative density.
The relative density (RD) is defined as the ratio of the density of a substance to the density of a reference substance. In this case, the reference substance is water (since the question does not specify another reference fluid).
1. Calculate the density of the solid:
Using the formula: Density = Mass / Volume
Density of the solid in air = Mass of the solid in air / Volume of the solid
Density of the solid in air = 200g / Volume of the solid
2. Calculate the density of the liquid:
Density of the liquid = Relative density x Density of the reference fluid
Relative density = 0.80 (given)
Density of the reference fluid (water) = 1000 kg/m³ (approximately)
3. Equate the densities of the solid and the liquid when the solid is partially immersed in the liquid:
Density of the solid in air = Density of the liquid
4. Solve for the Volume of the solid:
200g / Volume of the solid = 0.80 x 1000 kg/m³
Cross-multiplying:
200g = 0.80 x 1000 kg/m³ x Volume of the solid
200g = 800 kg/m³ x Volume of the solid
Dividing both sides by 800 kg/m³:
Volume of the solid = 200g / (800 kg/m³)
Volume of the solid = 0.25 m³
The volume of the solid immersed in the liquid is 0.25 m³.
To solve this problem, we need to use the principle of Archimedes' buoyancy.
First, we need to find the apparent loss of weight of the solid when it is immersed in the liquid. This can be calculated by subtracting the mass in the liquid from the mass in air:
Apparent Loss of Weight = Mass in Air - Mass in Liquid
Apparent Loss of Weight = 200g - 160g
Apparent Loss of Weight = 40g
According to Archimedes' principle, the apparent loss of weight is equal to the weight of the liquid displaced by the submerged portion of the solid.
Weight of liquid displaced = Apparent Loss of Weight
Weight of liquid displaced = 40g
Now, we can calculate the volume of the solid submerged in the liquid. We know that the relative density of the liquid is 0.80. Relative density is the ratio of the density of a substance to the density of a reference substance (usually water). In this case, it means that the density of the liquid is 0.80 times the density of water.
Density of water = Mass / Volume
Let V be the volume of the solid submerged in the liquid. The mass of the liquid displaced by this volume is equal to the density of the liquid times the volume:
Mass of liquid displaced = Density of liquid * Volume
Mass of liquid displaced = 0.80 * Mass of water * V
Since the weight of the liquid displaced is also equal to the apparent loss of weight of the solid, we can equate the two:
Weight of liquid displaced = Apparent Loss of Weight
0.80 * Mass of water * V = 40g
Now, we need to rearrange the equation to solve for V:
V = (Apparent Loss of Weight) / (0.80 * Mass of water)
Plugging in the values:
V = 40g / (0.80 * Mass of water)
To find the value of Mass of water, we need to consider the density of water, which is approximately 1g/cm³. Since density = mass/volume, we know that the mass of 1cm³ of water is 1g. So, replacing Mass of water by 1g:
V = 40g / (0.80 * 1g)
V = 50 cm³
Therefore, the volume of the solid immersed in the liquid is 50 cm³.